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PrudensOptimus

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My textbook failed to address how to find the epsilon and the delta of a Limit. Can somebody explain to me what they are and what they do?

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- Thread starter PrudensOptimus
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- #1

PrudensOptimus

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- #2

Welcome to the forums!

What class is this for? I have no idea what those words mean! Except Delta...

What class is this for? I have no idea what those words mean! Except Delta...

- #3

PrudensOptimus

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Calculus. Where else do you think they will teach Limits?

- #4

I may take calc in 1 variable as an elective fill-in, but more likely I'll take something else more usefull in my career.

- #5

Hurkyl

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Limits don't have an epsilon (ε) and a delta (δ) per se...

The idea of a limit is that it's supposed to capture our idea of "approaching" a number. For example, if I have the function:

f(x) = (x

I know that f(1) does not exist (because 0/0 is undefined). However, if I compute some values of f near 1:

f(1.1) = -1.9

f(1.01) = -1.99

f(1.001) = -1.999

et cetera

It appears to me that f(x) is approaching -2 as x approaches 1. Incidentally, we can write this in symbols as:

f(x) -> -2 as x -> 1

(where -> is supposed to be an arrow, not a dash followed by a greater than sign)

So the question is, how would we

Well, it's difficult to describe the idea of a moving point in mathematics, so instead let's just let x be some point near 1, and require that it has to be

Let the symbol δ be that variable. I shall require the distance between x and 1 to be less than δ. IOW:

|x - 1| < δ

Also, don't forget x isn't allowed to be exactly 1! Thus:

0 < |x - 1| < δ

Now, what about the value of f(x)? Let's use the same idea! Let's introduce the variable ε to be the bound on how close f(x) must be to -2. IOW, I want:

|f(x) - (-2)| < ε

Now, let us combine these two ideas. I want to say "When x is near 1, f(x) is near -2", so putting that into symbols:

I want to prove that: 0 < |x - 1| < δ implies |f(x) - (-2)| < ε

But we still have to figure out what &delta and &epsilon should be!

Well, let's defer that a little bit longer and figure out how they should be related. Our end goal is to prove that f(x) is approaching (-2), so let's fix ε and see if we can find a formula for δ that makes the implication true.

(note: These next steps are typical of so-called "ε-δ" proofs)

Goal: Fnid δ so that:

0 < |x - 1| < δ implies |f(x) - (-2)| < ε

First, plug in what f(x) is:

|(x

Multiply through by |x-1|:

|(x

|x

|(x - 1)

now divide

|x-1| < ε

And now it's clear how to choose our value for δ! If we simply choose δ to be equal to ε, then we're appear to be guaranteed that |x-1| < ε. However, the steps we did above are in the

Choose δ to be equal to ε

0 < |x - 1| < δ

substituting yields: |x - 1| < ε

Multiply by |x-1|: |x

Add and subtract inside the ||: |(x

Substitute f: |f(x) - (-2)| < ε

So we've proven our goal! For any ε we choose, we may select δ to be equal to ε and then the following is true:

0 < |x - 1| < δ implies |f(x) - (-2)| < ε

IOW, no matter how close we want f(x) to be to (-2), we are able to sufficiently narrow the freedom on x so that f(x) is close enough to (-2) for all allowed values of x.

But we still haven't chosen an actual value for ε! And this is the key to how the definition of a limit corresponds to our intuitive idea.

Anyways, that's the motivation for the whole thing. Summing it up in the rigorous definition:

f(x) -> L as x -> a if and only if

For every ε>0 there exists a δ>0 such that for all x:

0 < |x - a| < δ implies |f(x) - L| < ε

And to prove that a limit is a particular value using this definition, the typical strategy is what we did above; we "solve" for δ in terms of ε and then prove our solution yields the implication required by the definition.

And getting back to my original statement, δ and ε are

- #6

NEOclassic

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Originally posted by Hurkyl

Limits don't have an epsilon (ε) and a delta (δ) per se...

The idea of a limit is that it's supposed to capture our idea of "approaching" a number. For example, if I have the function:

f(x) = (x^{2}- 4x + 3) / (x - 1)

I know that f(1) does not exist (because 0/0 is undefined). However, if I compute some values of f near 1:

f(1.1) = -1.9

f(1.01) = -1.99

f(1.001) = -1.999

et cetera

Hi Hurkyl,

Also: f(.9) = -2.1

f(.99) = -2.01

f(.999) = -2.001 et cetera

and I learned in college that the validation of this technique was referred to as "Cauchy Critereon" - "For every neighborhood, no matter how small, - - -etc etc."

Of course there is no argument that the limit of the example is really - 2; and when I tried to quick fix using l'Hospital's treatment, your arduous but elegantly explicit post was verified post haste.

The Rule states that if the ratio of two functions involves division by zero at some limiting value, the ratio of the derivatives of those functions (or 2nd,3rd etc derivatives) may indicate the existence of a limit and, as I later learned, the actual value of that limit.

It's easy to see how it works for the limit (as x ->0) of [sin x]/x which is [cos x]dx/dx =1 Q.E.D.

similarly for your example the derivative ratio is [2x -4]dx/dx which when evaluated at x=1 gives [2x1 - 4] = -2 Q.E.D. Cheers, Jim

- #7

Hurkyl

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As NEO mentioned, there are other ways to

- #8

newton1

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Originally posted by NEOclassic

Hi Hurkyl,

Also: f(.9) = -2.1

f(.99) = -2.01

f(.999) = -2.001 et cetera

and I learned in college that the validation of this technique was referred to as "Cauchy Critereon" - "For every neighborhood, no matter how small, - - -etc etc."

Of course there is no argument that the limit of the example is really - 2; and when I tried to quick fix using l'Hospital's treatment, your arduous but elegantly explicit post was verified post haste.

The Rule states that if the ratio of two functions involves division by zero at some limiting value, the ratio of the derivatives of those functions (or 2nd,3rd etc derivatives) may indicate the existence of a limit and, as I later learned, the actual value of that limit.

It's easy to see how it works for the limit (as x ->0) of [sin x]/x which is [cos x]dx/dx =1 Q.E.D.

similarly for your example the derivative ratio is [2x -4]dx/dx which when evaluated at x=1 gives [2x1 - 4] = -2 Q.E.D. Cheers, Jim

using the epsilon (£`) and a delta (£_) is to prove the limit is logic

just like to prove limit x( x approach to 3 )= 3

although you know the answer is 3

but for the mathematics logic

you should show the equation is logic

and epsilon (£`) and a delta (£_) is the way

just like what hurkyl show

this are not using to find the answer of limet !

- #9

KLscilevothma

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"For every ε>0 there exists a δ>0 such that for all x:

0 < |x - a| < δ implies |f(x) - L| <ε " isn't true)

Since we are working backward to find the values of ε

and δ, I think it is harder to do the proof, right? Would you please give me an example using epsilon-delta argument to prove a limit doesn't exist? I've never met one in class.

Thanks

- #10

newton1

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Originally posted by KL Kam

"For every ε>0 there exists a δ>0 such that for all x:

0 < |x - a| < δ implies |f(x) - L| <ε " isn't true)

Since we are working backward to find the values of ε

and δ, I think it is harder to do the proof, right? Would you please give me an example using epsilon-delta argument to prove a limit doesn't exist? I've never met one in class.

Thanks

what you mean the limit does not exist??

did you mean the limit go to infinity? or discontinuity??

i think this can help...

for the infinite large definition

For each M>0 there exists a £_>0 such that

0 < |x - a| < £_ implies f(x) >M

infinite small definition is

For each M>0 there exists a £_>0 such that

0 < |x - a| < £_ implies f(x) <-M

and continuity definition is

For each £`>0 there exists a £_>0 such that

0 < |x - a| < £_ then |f(x) - f(a)| <£`

if the limit can not find some £_ in this definition

that mean the function discontinuity at a

- #11

newton1

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Originally posted by Hurkyl

As NEO mentioned, there are other ways tocomputelimits. It is important to learn the definition of a limit so you understand what the mathematics is trying to say (and so you can get that problem right on the test), but for practical computation, different techniques are used, which you will learn in a few sections.

wrong! Hurkyl...you wrong!

using the epsilon (£`) and a delta (£_) is to prove the limit is logic

just like to prove limit x( x approach to 3 )= 3

although you know the answer is 3

but for the mathematics logic

you should show the equation is logic

just like the example you write

you already know the answer of the limit

and you just prove the equation is logic

l'Hospital's rule is use to find the answer of the limit

this two skill are different...

epsilon (£`) and a delta (£_) can't do what l'Hospital rule do

and L'Hospital rule can't do what epsilon (£`) and a delta (£_) do

you are not grasp the meaning of epsilon (£`) and a delta (£_)

- #12

Hurkyl

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For instance, for my example problem, we had

f(x) = (x

However, notice the numerator can be factored into (x - 1) * (x - 3). If we let

g(x) = x - 3

then g(x) = f(x) for all x [x=] 1. We can prove, using the ε-δ definition the theorem that:

If f(x) = g(x) for all x such that p < x < a and a < x < q (for some p and q)

and f(x) -> L as x -> a

then g(x) -> L as x -> a

So then to find the limit of f(x) as x approaches 1, we simply need to find the limit of g(x) as x approaches 1.

We can also prove the following theorem, again using the ε-δ definition:

If h(x) is a polynomial, then h(x) -> h(a) as x -> a

So then we know that g(x) -> g(1) as x -> 1.

And we combine the two theorems to prove

f(x) -> -2 as x -> 1

- #13

newton1

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Originally posted by Hurkyl

For instance, for my example problem, we had

f(x) = (x^{2}- 4x + 3) / (x - 1)

However, notice the numerator can be factored into (x - 1) * (x - 3). If we let

g(x) = x - 3

then g(x) = f(x) for all x [x=] 1. We can prove, using the ε-δ definition the theorem that:

If f(x) = g(x) for all x such that p < x < a and a < x < q (for some p and q)

and f(x) -> L as x -> a

then g(x) -> L as x -> a

So then to find the limit of f(x) as x approaches 1, we simply need to find the limit of g(x) as x approaches 1.

We can also prove the following theorem, again using the ε-δ definition:

If h(x) is a polynomial, then h(x) -> h(a) as x -> a

So then we know that g(x) -> g(1) as x -> 1.

And we combine the two theorems to prove

f(x) -> -2 as x -> 1

no...no...

first...you already know the answer of the g(x)

second , if f(x) is unknown

how can you assume f(x)--> a and g(x)--> a

if f(x) is so complex

how can you find some function g(x) is eqaul to f(x)

you assume the f(x)--> a and g(x)--> a

because you already know the answer of f(x)

if not, you will not set the g(1)--> 2

i told you already £`-£_ definition just to prove the limit is logic when you already know the answer

L'Hospital rule is find the answer of the unknown limit

but £`-£_ definition can't do that

- #14

KLscilevothma

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Originally posted by Newton1

no...no...

first...you already know the answer of the g(x)

second , if f(x) is unknown

how can you assume f(x)--> a and g(x)--> a

According to Hurkyl's last post, he didn't know the limit value of both f(x) and g(x) at the very beginning. He only knew f(x) can be simplified to g(x), where x [x=] 1, after factorization.

He didn't set limit value of g(x) tends to 2 as x tends to 1.you assume the f(x)--> a and g(x)--> a

because you already know the answer of f(x)

if not, you will not set the g(1)--> 2

He used epsilon-delta (e-d) defination to prove the following :

If f(x) = g(x) for all x such that p < x < a and a < x < q (for some p and q) and f(x) -> L as x -> a

then g(x) -> L as x -> a ...(1)

In his example, f(x)=g(x) for all x such that p<x<1 and 1<x<q for some p and q and f(x) -> L as x ->a. (We still don't know the the limit valye, L up to now)

Then he proved, using e-d defination that

If h(x) is a polynomial, then h(x) -> h(a) as x -> a ... (2)

By combining (1) and (2), we know that we can substitute x=1 into g(x) to get the limit value of f(x)

We don't know the limit value of f(x) before step (1) and (2) but we've used e-d defination to show that we can use (1) and (2) to compute the limit value of f(x)£`-£_ definition just to prove the limit is logic when you already know the answer

L'Hospital rule is find the answer of the unknown limit

but £`-£_ definition can't do that

I think the L'hospital rule is somehow based on the e-d defination, but not the other way round.

- #15

newton1

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Originally posted by KL Kam

According to Hurkyl's last post, he didn't know the limit value of both f(x) and g(x) at the very beginning. He only knew f(x) can be simplified to g(x), where x [x=] 1, after factorization.

He didn't set limit value of g(x) tends to 2 as x tends to 1.

He used epsilon-delta (e-d) defination to prove the following :

If f(x) = g(x) for all x such that p < x < a and a < x < q (for some p and q) and f(x) -> L as x -> a

then g(x) -> L as x -> a ...(1)

In his example, f(x)=g(x) for all x such that p<x<1 and 1<x<q for some p and q and f(x) -> L as x ->a. (We still don't know the the limit valye, L up to now)

Then he proved, using e-d defination that

If h(x) is a polynomial, then h(x) -> h(a) as x -> a ... (2)

By combining (1) and (2), we know that we can substitute x=1 into g(x) to get the limit value of f(x)

We don't know the limit value of f(x) before step (1) and (2) but we've used e-d defination to show that we can use (1) and (2) to compute the limit value of f(x)

I think the L'hospital rule is somehow based on the e-d defination, but not the other way round.

ok

now i give you a question

then you try to use the £`-£_ definition to find the answer

but...don't use the L'Hospital rule to find the answer

ok??

limit [(1/x)-1/((e^x)-1)]=?? ...(x-->0)

if you can do this with £`-£_ definition

show you step

and...don't use the other way to find the answer 1st!

- #16

Hurkyl

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ok

now i give you a question

then you try to use the £`-£_ definition to find the answer

but...don't use the L'Hospital rule to find the answer

ok??

limit [(1/x)-1/((e^x)-1)]=?? ...(x-->0)

if you can do this with £`-£_ definition

show you step

and...don't use the other way to find the answer 1st!

I'm not sure what point you're trying to make by this.

I could just use the ε-&delta method to prove L'Hopital's rule and use L'Hopital's rule to compute the limit, and I would be fully within the constraints of your challenge because I did the whole thing using the ε-δ definition.

That's why we use things like L'Hopital's rule to evaluate limits in practice; they are rigorously proven in terms of the ε-δ definition, yet are simpler to apply... but since we've proven L'Hopital's rule in terms of the ε-δ definition, we can, in principle, just cut-paste the proof of L'Hopital's rule (and other theorems) into our proofs to give a proof that uses nothing but ε-δ, though it could be very long and complex. We use theorems in mathematics so we don't have to cut and paste.

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- #17

newton1

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Originally posted by Hurkyl

I'm not sure what point you're trying to make by this.

I could just use the ε-&delta method to prove L'Hopital's rule and use L'Hopital's rule to compute the limit, and I would be fully within the constraints of your challenge because I did the whole thing using the ε-δ definition.

just like what i say before

£`-£_ definition can't use to find the answer of limit

this definition just use to proof your answer is right

but L'Hospital just only can find the answer

this 2 skill is different

but your people say £`-£_ is "the other ways to compute limits"

this is absolutely wrong!

- #18

Hurkyl

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Use of L'Hopital's rule is most certainly *proof* that the limit has a particular value.

- #19

newton1

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Originally posted by Hurkyl

Use of L'Hopital's rule is most certainlyproofthat the limit has a particular value.

i am lazy to say again

you really don't know what differeent with "prove it exist" and "find the value"

- #20

NEOclassic

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Originally posted by Hurkyl

Use of L'Hopital's rule is most certainlyproofthat the limit has a particular value.

Hi again Hurkyl,

When I check the profiles of this string, I see that posters are students (high-school and Bacc)and because I learned about the power of l'Hospital's rule and Cauchy Criterion in graduate studies, I submit that the adversarial nature of this string is likely caused by the absence of a complete deck. Challenge is often the route to learning that must be tolerated and that is okay.

Clearly here, some of these students have possibly learned something new (the rule and the criterion) from our postings and I propose that they check them out at the school library or Google.

I have discovered that if the derivative of a function divides the function itself, the result is the Newton-Raphson root-extraction method (see Greg's other forum).

Newton1 will have trouble when he tries to understand how some stepwise functions have dual limits dependent on whether analysis is from the left or the right; the "proof of the pudding is - -" and all that rot! Cheers, Jim PS: Thanks for your patience with these kids.

- #21

Hurkyl

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i am lazy to say again

you really don't know what differeent with "prove it exist" and "find the value"

Well, maybe you can clarify specifically what you mean by those terms.

I'm incredulous at your assertion L'Hopital's rule can do at least one thing you can't do with an ε-δ method, because L'Hopital's rule is a deduction from the ε-δ method, and anytime you use L'Hopital's rule for anything, you could just directly substitute the ε-δ derivation so that you get the same results, but your derivation is solely in terms of ε-δ and L'Hopital's rule never appears in the problem.

- #22

Integral

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So, if you have an expression which satisfies the existence conditons of the proof of a limit you can be sure that the limit exists. But it does not tell you how to determine what that limit is. That is a different matter altogether. Various tools, such as L'Hoptls rule have been developed to aid in that process. The only wat to determine if a given function or series has a limit it to know how to apply the tools available and to know how to algebraiaicly manipulate the expression to perserve the limit at the same time you expose it.

This is one of those things any 5 yr onld can do..

With 10yrs experiance.

- #23

newton1

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Originally posted by Integral

So, if you have an expression which satisfies the existence conditons of the proof of a limit you can be sure that the limit exists. But it does not tell you how to determine what that limit is. That is a different matter altogether. Various tools, such as L'Hoptls rule have been developed to aid in that process. The only wat to determine if a given function or series has a limit it to know how to apply the tools available and to know how to algebraiaicly manipulate the expression to perserve the limit at the same time you expose it.

This is one of those things any 5 yr onld can do..

With 10yrs experiance.

exactly!

- #24

I think the whole argument of limit started with Zeno’s paradox of motion. In this paradox he essentially argued that motion is an illusion of mind since it is impossible to move from point A to point B . To move from point A to point B one has to go through infinite number of points therefore such a motions would take infinite time to be completed. For example, to go from number 1 to number 2 there are infinite numbers of points like 1.2, 1.5, 1.9… that one has to go through, so how number 1 ever reaches the number 2?

To answer this paradox many philosophers and mathematicians brought forth several rhetorical and logical explanations and the limit definition is, I think, the result.

- #25

The definition of limit can work to explain and resolve paradoxes such as Zeno’s by incorporating logic using mathematical denotations.

For example if two points of A and B are placed on two-dimensional rectangular coordination we can incorporate this definition by approaching the problem as it is defined.

We want to place point A so close to point B that that their distance in x and y direction be minuscule but not zero. The coordination of these two points is as follows:

Point A: (x, f (x)) and point B: (x->c, f (x->c)=L)

We want these points to be as close as possible so their absolute distances of |x-c| in x direction and |f (x)-L| in y direction to be as close to zero as possible but never zero.

0 < |x-c| < δ and |f (x)-L| < ε

- #26

We see that f (1.999)≈ 3.996 and f (2.001)≈ 4.004. So |f (x) –L| is 4.004-4= .004 and 3.996-4=-.004 hence ε= |. 004|=. 004 or ε=4σ

We see that as we approach point B from its left and from its right point B approaches a Limit number L from both directions then the limit exists and point B can be reached.

This is like saying that number 2 between number 1 and number 3 exists because if one approaches number 2 from number 1 (left side) or from number 3 (right side) one still reaches number 2.

- #27

NEOclassic

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Hello String,

While in graduate school and engulfed in Kaplan's "Advanced Calculus Theory" I ran into a function that may have been unique because it occurred for only one integer exponent in the set of so-called "power series" of the form: f(x) = k / x^n where n is an integer. As I recall, at all values of "n" the limit as x -> 0 is asymptotic to the y axis; however, for at least one integer the area under the curve between zero and any x is finite.

This is really not show boating but rather a challenge for you calculus students to ponder over. I believe that the integral dx/x = ln x has somthing to do with understanding this mystery. Cheers, Jim

- #28

Hurkyl

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only exists for n > -1, so what you're remembereing must be something different.. I can't think of what that might be.

PS: For those interested, the series:

(Σ

is called a Laurent series.

- #29

NEOclassic

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I really don't remember the exact function but the detailed discussion that occurred in class was real. I'm pretty sure that we were in the textbook area having to do with powers of the independent variable in the denominator. My Kaplan disappeared at least 35 years ago and I'm too busy to browse the two or so cubic feet of notes that I think are somewhere in a box. I thank you for calling out my mis-thought. Maybe I'll take time to browse my distant past later. Cheers, Jim

- #30

Hurkyl

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&int

- #31

NEOclassic

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Hi Hurkyl,

By Jove,I think you've got it! Cheers, Jim

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